A Paradox of the Mean Variance Setting for the Long Term Investor

Abstract

We show that the mean-variance preferences have counterfactual implications for a risk averse long term decision maker. In the simple case of dynamic portfolio choice, we show that the optimal certainty equivalent is decreasing with the investor’s horizon towards its lower bound, the riskless rate. For some horizons (less than 25 years in our simulations), the economic value of diversification is 0 and therefore the optimal portfolio strategy is a buy and hold one in the riskless asset. Therefore, under-diversification is optimal. These results question the usefulness of the mean variance setting for long term dynamic decision making.

Appendix

Proof of Proposition 1:

Investing in the risky assets is optimal if the following condition hold:

$$\begin{aligned} \mathrm{{ce}}_\mathrm{{T}}^{*} \ge -\frac{1}{\mathrm{{T}}}\ln \mathrm{{P}}\left( {0, \mathrm{{T}}} \right) \end{aligned}$$

(A1)

As a consequence, we have:

$$\begin{aligned} \frac{1}{\mathrm{{T}}}\ln \left[ {\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }-\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2} \right] ^{-\frac{1}{2}}} \right] \ge -\frac{1}{\mathrm{{T}}}\ln P\left( {0, \mathrm{{T}}}\right) \end{aligned}$$

(A2)

$$\begin{aligned} \Leftrightarrow \mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}\le \frac{1}{\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) } \end{aligned}$$

On the other hand, we have:

$$\begin{aligned}&\frac{\partial \mathrm{{ce}}_\mathrm{{T}}^{*} }{\partial \mathrm{{T}}} = -\frac{1}{\mathrm{{T}}^{2}}\ln \left[ {\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }-\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}} \right] \\&\qquad \qquad \qquad +\frac{1}{\mathrm{{T}}}\frac{1}{\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0} -\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}}\frac{\partial }{\partial \mathrm{{T}}}\left[ {\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }-\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0}{\mathrm{{W}}_0}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}} \right] \end{aligned}$$

(A3)

A sufficient condition for this derivative to be negative is:

$$\begin{aligned} \frac{\partial }{\partial \mathrm{{T}}}\left[ {\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }-\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}} \right] <0 \end{aligned}$$

(A4)

Since we have:

$$\begin{aligned} \frac{\partial }{\partial \mathrm{{T}}}\left[ {\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }-\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}} \right]&=-\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) \\&\qquad \qquad -\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0,\mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^\frac{1}{2} \end{aligned}$$

(A5)

Condition (A5) holds whenever:

$$\begin{aligned}&-\frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\frac{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }{\mathrm{{W}}_0 }\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}<0\\&\qquad \Leftrightarrow \frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}>\frac{\frac{1}{\upgamma }}{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}\left[ {-\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) } \right] \end{aligned}$$

(A6)

Using (A2), condition (A6) becomes:

$$\begin{aligned}&\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}>\frac{\frac{1}{\upgamma }}{\frac{1}{\upgamma }\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\mathrm{{W}}_0 }\frac{1}{\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) }\left[ {-\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) } \right] \\&\qquad \Leftrightarrow \frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}>\frac{1}{\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) -\frac{\mathrm{{W}}_0 }{\frac{1}{\upgamma }}}\frac{1}{\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) }\left[ {-\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) } \right] \end{aligned}$$

(A7)

Condition (A7) holds when

$$\begin{aligned} \frac{\partial }{\partial \mathrm{{T}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{T}}^2 } \right] ^{-\frac{1}{2}}>\frac{1}{\mathrm{{P}}\left( {0,\mathrm{{ T}}} \right) -1}\frac{1}{\mathrm{{P}}\left( {0,\mathrm{{ T}}} \right) }\left[ {-\frac{\partial }{\partial \mathrm{{T}}}\mathrm{{P}}\left( {0, \mathrm{{T}}} \right) } \right] \end{aligned}$$

(A8)

holds and this will always be the case since the right hand side is always negative while the left hand side is always positive.

Proof of Proposition 2:

The first order condition writes:

$$\begin{aligned} \delta ^{\mathrm{{t}}}\left( {1-\upgamma \mathrm{{c_t}}^{*} } \right) -\uptheta \Lambda _\mathrm{{t}} =0\Leftrightarrow \mathrm{{c_t}}^{*} =\frac{1}{\upgamma }\left( {1-\uptheta \delta ^{-\mathrm{{t}}}\Lambda _\mathrm{{t}}} \right) \end{aligned}$$

(A9)

where $\uptheta$ is the Lagrange multiplier. Using the budget constraint, the Lagrange multiplier is given by:

$$\begin{aligned} \mathrm{{E}}_0 \left[ \int _0^\mathrm{{T}} \frac{1}{\upgamma }\left( 1-\uptheta \delta ^{-\mathrm{{t}}}\Lambda _\mathrm{{t}} \right) \mathrm{{dt}} \right] =\mathrm{{W}}_0 \Leftrightarrow \frac{\frac{1}{\upgamma }\mathrm{{E}}_0 \left[ \int \nolimits _0^\mathrm{{T}} \Lambda _\mathrm{{t}}\mathrm{{dt}} \right] - \mathrm{{W}}_0}{\frac{1}{\upgamma }\mathrm{{E}}_0 \left[ \int _0^\mathrm{{T}} \delta ^{-\mathrm{{t}}}\Lambda _\mathrm{{t}}^2 \mathrm{{dt}} \right] }=\uptheta \end{aligned}$$

(A10)

This could be written as:

$$\begin{aligned} \uptheta =\frac{\frac{1}{\upgamma }\int _0^\mathrm{{T}}{\mathrm{{P}}\left( {0,\mathrm{{t}}} \right) \mathrm{{dt}}} \mathrm{{W}}_0 }{\frac{1}{\upgamma }\int _0^\mathrm{{T}} {\delta ^{\mathrm{{-t}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{t}}^2 } \right] \mathrm{{dt}}} } \end{aligned}$$

(A11)

It has a similar interpretation as the one obtained in (4). It is positive when the present value of all maximal consumptions is greater than initial wealth. The Dollar certainty equivalent for consumption at time t is such that:

$$\begin{aligned} \mathrm{{E}}_0 \left[ {\mathrm{{c_t}}^{*} -\frac{\upgamma }{2}\left( {\mathrm{{c_t}}^{*} } \right) ^{2}} \right] =\mathrm{{CE_t}}^{*} -\frac{\upgamma }{2}\left( {\mathrm{{CE_t}}^{*} } \right) ^{2} \end{aligned}$$

(A12)

Therefore, using the expression for the optimal consumption, one has:

$$\begin{aligned} \mathrm{{CE_t}}^{*} =\frac{1}{\upgamma }\left[ {1-\uptheta \delta ^{-\mathrm{{t}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{t}}^2 } \right] ^{-\frac{1}{2}}} \right] \end{aligned}$$

(A13)

The certainty equivalent associated with the optimal consumption plan will thus be:

$$\begin{aligned} \mathrm{{CE}}_{0,\mathrm{{T}}}^{*}=\int _{0}^{\mathrm{{T}}}\frac{\mathrm{{P(0,t)}}}{\mathrm{{P(0,T)}}}\mathrm{{CE}}^{*}_{\mathrm{{t}}}\mathrm{{dt}} \end{aligned}$$

(A14)

and therefore:

$$\begin{aligned} \mathrm{{CE}}^{*}_{0,\mathrm{{T}}}=\frac{1}{\upgamma }\int _{0}^{\mathrm{{T}}}\frac{\mathrm{{P(0,t)}}}{\mathrm{{P(0,T)}}}\left[ 1-\uptheta \delta ^\mathrm{{-t}}\mathrm{{E}}_{0}[\nabla _\mathrm{{t}}^{2}]^{1/2}\right] \mathrm{{dt}} \end{aligned}$$

(A15)

The annualized certainty equivalent becomes:

$$\begin{aligned} \mathrm{{ce}}_{0,\mathrm{{T}}}^{*} =&\frac{1}{\mathrm{{T}}}\ln \frac{\mathrm{{CE}}_{0,\mathrm{{T}}}^{*}}{\mathrm{{W}}_0 } \nonumber \\ =&\frac{1}{\mathrm{{T}}}\ln \frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0}+\frac{1}{\mathrm{{T}}}\ln \left[ \int \nolimits _0^\mathrm{{T}} \frac{\mathrm{{P}}( 0, \mathrm{{t}} )}{\mathrm{{P}}(0, \mathrm{{T}})} \left[ 1-\uptheta \delta ^{-\mathrm{{t}}}\mathrm{{E}}_0 \left[ \Lambda _\mathrm{{t}}^{2} \right] ^{-\frac{1}{2}} \right] \mathrm{{dt}}\right] \end{aligned}$$

(A16)

Substituting for the Lagrange multiplier yields:

$$\begin{aligned} \mathrm{{ce}}_{0, \mathrm{{T}}}^{*}&= \frac{1}{\mathrm{{T}}}\ln \frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 } \nonumber \\&+\frac{1}{\mathrm{{T}}}\ln \left[ \int \nolimits _0^\mathrm{{T}} \frac{\mathrm{{P}}( 0, \mathrm{{T}})}{\mathrm{{P}}( 0, \mathrm{{T}} )} \mathrm{{dt}}-\frac{\frac{1}{\upgamma }\int _0^\mathrm{{T}} \mathrm{{P}}( 0, \mathrm{{T}})\mathrm{{dt}} -\mathrm{{ W}}_0}{\frac{1}{\upgamma }\int _0^\mathrm{{T}} \delta ^{-t}\mathrm{{E}}_0 \left[ \Lambda _\mathrm{{t}}^2 \right] \mathrm{{dt}}}\int \nolimits _0^\mathrm{{T}} \frac{\mathrm{{P}}(0,\mathrm{{t}})}{0,\mathrm{{T}}}\delta ^{-\mathrm{{t}}}\mathrm{{E}}_0 \left[ \Lambda _\mathrm{{t}}^{2} \right] ^{-\frac{1}{2}}\mathrm{{dt}}\right] \end{aligned}$$

(A17)

Under the specific setting outlined above, this becomes:

$$\begin{aligned} \mathrm{{ce}}_{0, \mathrm{{T}}}^{*} =\;&\frac{1}{\mathrm{{T}}}\ln \frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }\nonumber \\&+\frac{1}{\mathrm{{T}}}\ln \left[ \frac{1}{\mathrm{{r}}}\left( {\mathrm{{e}}^{\mathrm{{rT}}}-1} \right) -\frac{\ln \delta +2\left( {\mathrm{{r}}-\frac{\kappa {\prime }\kappa }{2}} \right) }{\ln \delta +\frac{\kappa {\prime }\kappa }{2}}\frac{\mathrm{{e}}^{-\left( {\ln \delta +\frac{\kappa {\prime }\kappa }{2}} \right) \mathrm{{T}}}-1}{\mathrm{{e}}^{-\left( {\ln \delta +2\left( {\mathrm{{r}}-\frac{\kappa {\prime }\kappa }{2}} \right) } \right) \mathrm{{T}}}-1}\right. \\&{}\times \left. \left[ {-\frac{1}{\mathrm{{r}}}\left( {1-\mathrm{{e}}^{\mathrm{{rT}}}} \right) -\frac{\mathrm{{W}}_0 \mathrm{{e}}^{\mathrm{{rT}}}}{\frac{1}{\upgamma }}} \right] \right] \end{aligned}$$

(A18)

Note also that in the particular case of constant opportunity set, the condition (A11) implies:

$$\begin{aligned} \frac{\frac{1}{\upgamma }\int _0^\mathrm{{T}} {\mathrm{{P}}\left( {0,\mathrm{{t}}} \right) \mathrm{{dt}}}- \mathrm{{W}}_0}{\frac{1}{\upgamma }\int _0^\mathrm{{T}} {\delta ^{-\mathrm{{t}}}\mathrm{{E}}_0 \left[ {\Lambda _\mathrm{{t}}^2 } \right] \mathrm{{dt}}} }>0\Leftrightarrow \frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }>\frac{\mathrm{{r}}}{1-\mathrm{{e}}^{-\mathrm{{rT}}}} \end{aligned}$$

(A19)

The maximum certainty equivalent is the one that would obtain when consumption is equal to the upper bound at each time t in the future. Therefore:

$$\begin{aligned} \mathrm{{ce}}_{0, \mathrm{{T}}}^{\max } =\frac{1}{\mathrm{{T}}}\ln \left[ \frac{\frac{1}{\upgamma }}{\mathrm{{W}}_0 }\int \nolimits _0^\mathrm{{T}} \frac{\mathrm{{P}}( 0, \mathrm{{t}} )}{\mathrm{{P}}( 0, \mathrm{{T}} )} \mathrm{{dt}} \right] \end{aligned}$$

(A20)